a: Ta có: \(\dfrac{1}{9}\cdot27^x=3^x\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^x=\dfrac{1}{9}\)
hay x=1
b: Ta có: \(\left(3x+4\right)^2-8=41\)
\(\Leftrightarrow\left(3x+4\right)^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=7\\3x+4=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\3x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{3}\end{matrix}\right.\)
c: Ta có: \(\dfrac{\left(-3\right)^x}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4\)
Suy ra: x=7
a: Ta có: \(\dfrac{1}{9}\cdot3^4\cdot3^x=3^x\)
\(\Leftrightarrow3^{x+2}-3^x=0\)
\(\Leftrightarrow x+2=x\left(loại\right)\)
b: Ta có: \(2\cdot\left(x+\dfrac{1}{4}\right)^3=-\dfrac{27}{4}\)
\(\Leftrightarrow x+\dfrac{1}{4}=-\dfrac{3}{2}\)
hay \(x=-\dfrac{7}{4}\)
c: Ta có: \(x:\left(-\dfrac{1}{3}\right)^2=-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{1}{3}\cdot\dfrac{1}{9}\)
hay \(x=-\dfrac{1}{27}\)
a: Ta có: \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)
\(\Leftrightarrow3^{2x}=\dfrac{1}{9}\)
\(\Leftrightarrow2x=-2\)
hay x=-1
b: Ta có: \(\dfrac{1}{4}+\left(2x-1\right)^3=\dfrac{1}{8}\)
\(\Leftrightarrow2x-1=-\dfrac{1}{2}\)
hay \(x=\dfrac{1}{4}\)
